Let $x,$ $y,$ $z$ be nonnegative real numbers.  Let
\begin{align*}
A &= \sqrt{x + 2} + \sqrt{y + 5} + \sqrt{z + 10}, \\
B &= \sqrt{x + 1} + \sqrt{y + 1} + \sqrt{z + 1}.
\end{align*}Find the minimum value of $A^2 - B^2.$
Answer: We can write
\begin{align*}
A^2 - B^2 &= (A + B)(A - B) \\
&= (\sqrt{x + 2} + \sqrt{x + 1} + \sqrt{y + 5} + \sqrt{y + 1} + \sqrt{z + 10} + \sqrt{z + 1}) \\
&\quad \times (\sqrt{x + 2} - \sqrt{x + 1} + \sqrt{y + 5} - \sqrt{y + 1} + \sqrt{z + 10} - \sqrt{z + 1}).
\end{align*}Let
\begin{align*}
a_1 &= \sqrt{x + 2} + \sqrt{x + 1}, \\
b_1 &= \sqrt{y + 5} + \sqrt{y + 1}, \\
c_1 &= \sqrt{z + 10} + \sqrt{z + 1}, \\
a_2 &= \sqrt{x + 2} - \sqrt{x + 1}, \\
b_2 &= \sqrt{y + 5} - \sqrt{y + 1}, \\
c_2 &= \sqrt{z + 10} - \sqrt{z + 1}.
\end{align*}Then by Cauchy-Schwarz,
\begin{align*}
A^2 - B^2 &= (a_1 + b_1 + c_1)(a_2 + b_2 + c_2) \\
&\ge (\sqrt{a_1 a_2} + \sqrt{b_1 b_2} + \sqrt{c_2 c_2})^2 \\
&= (1 + 2 + 3)^2 \\
&= 36.
\end{align*}Equality occurs when
\[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2},\]or equivalently,
\[\frac{x + 2}{x + 1} = \frac{y + 5}{y + 1} = \frac{z + 10}{z + 1}.\]For example, if we set each fraction to 2, then we get $x = 0,$ $y = 3,$ and $z = 8.$

Hence, the minimum value is $\boxed{36}.$